The formula answers: how many tuples $(\sigma_1,\sigma_2,...,\sigma_n)$ of elements of a given group G such that
(1) $\sigma_i\in C_i$ , where $C_i$ stands for conjugacy class.
(2) $\sigma_1\sigma_2...\sigma_n= id$

I want to know the name and exact content of this formula. Also , is there any connection between this formula and Burnside counting theorem (orbit-counting theorem)? I also want of a proof of the formula using idompotent (perhaps) and other related theorems.

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I wanna give a ppt on representaiton theorem .My lecturer mentioned this result and I wanna include it in the talk. guess each C-i should stand for different conjugacy class.
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user 566Dec 31 '12 at 13:09

Sorry I deleted my earlier comment as it seemed irrelevant after you got the answer on MSE. Thanks for your reply.
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Michael MurrayDec 31 '12 at 13:29

2 Answers
2

I just answered this on mathstackexchange, but I duplicate it here (later edit: But include here a proof for the more general formula):
You haven't told the reader what the formula is. I believe I know the formula you mean,
which is $$ \left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(\sigma_{j})|} \right)
\sum_{i=1}^{k} \frac{\prod_{j = 1}^{n} \chi_{i}(\sigma_{j})}{\chi_{i}(1)^{n-2}},$$ where the $\chi_{i}$ are all the complex irreducible characters of $G$. This formula was probably known to Burnside- I know no special name for it- it is a special case of a general formula for the product of class sums in the group algebra. The formula can be derived by writing the class sums as linear combinations of the primitive idempotents of $Z(\mathbb{C}G).$ Such linear combinations are easy to multiply since these idempotents are mutually orthogonal. Then one recovers the coefficient of a particular element $g$ in the product by using the fact that $g$ occurs with coefficient $\frac{\chi(1)\overline{\chi(g)}}{|G|}$ in the primitive central idempotent corresponding to the irreducible character $\chi$. The formula for the product of two class sums appears explicitly in Burnside's book, but some of the exercises in the book make it clear that he was aware of the general formula. As far as I know, there is no obvious connection with Burnside's orbit counting formula.

Later edit in response: I outlined above how to obtain the general formula, which eally indicates the proof: Let $e_{i}$ be the pirmitive central idempotent corresponding to the irreducible character $\chi_{i}.$ Then $e_{i} = \frac{\chi_{i}(1)}{|G|} \sum_{g \in G}\chi_{i}(g^{-1})g.$ Let $x_{j}$ be a representative for the $j$-the conjugacy class $C_{j}.$ Let $C_{j}^{+}$ denote the class sum of $C_{j}$ in the center of the group algebra. Then evaluating the central characters on both sides shows that
$C_{j}^{+} = \sum_{i=1}^{k} \frac{[G:C_{G}(x_{j})]\chi_{i}(x_{j})}{\chi_{i}(1)} e_{i}.$

Now if we take any $n$ class sums, which we may as well label $C_{1}^{+},\ldots, C_{n}^{+},$ we multiply them using the orthogonality of the central primitive idempotents,
and then recover the coefficient of an element $x_{n+1}$ in the product, to see that the coefficient of $x_{n+1}$ in $\prod_{j=1}C_{j}^{+}$ is
$$ \left( \frac{|G|^{n-1}}{\prod_{j=1}^{n} |C_{G}(x_{j})|} \right)
\sum_{i=1}^{k} \frac{\chi_{i}(x_{n+1}^{-1})\prod_{j = 1}^{n} \chi_{i}(x_{j})}{\chi_{i}(1)^{n-1}}.$$

Thanks .I wanna read a complete or good proof of the formula . Is there any online reference ? Also, you said it's a special case of a general formula . What is this general formula?
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user 566Dec 31 '12 at 13:32

Hvae now added a pretty complete proof of the general formula
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Geoff RobinsonDec 31 '12 at 14:14

The (general) formula keeps occurring in the literature from time to time. It was used in the 1980s by J.G. Thompson and others in connection with realizing groups as Galois groups- to establish a property called rigidity.
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Geoff RobinsonDec 31 '12 at 14:36